3-2/9b=1/3b-7=b

Solution for 3-2/9b=1/3b-7=b equation:

3-2/9b=1/3b-7=b

We move all terms to the left:

3-2/9b-(1/3b-7)=0


Domain of the equation: 9b!=0

b!=0/9

b!=0

b∈R



Domain of the equation: 3b-7)!=0

b∈R


We get rid of parentheses

-2/9b-1/3b+7+3=0

We calculate fractions


(-6b)/27b^2+(-9b)/27b^2+7+3=0

We add all the numbers together, and all the variables

(-6b)/27b^2+(-9b)/27b^2+10=0

We multiply all the terms by the denominator


(-6b)+(-9b)+10*27b^2=0

Wy multiply elements

270b^2+(-6b)+(-9b)=0

We get rid of parentheses

270b^2-6b-9b=0

We add all the numbers together, and all the variables

270b^2-15b=0

a = 270; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·270·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*270}=\frac{0}{540}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*270}=\frac{30}{540}
=1/18
$